logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Find the integrals of the functions\[\sin^2(2x+5)\]

Can you answer this question?
 
 

1 Answer

0 votes

Solution:

$\frac{1}{2}x - \frac{1}{8}sin(4x+10)+k$


Toolbox
  • Anytime you see a problem that is has a $\sin^2x$ function, you cannot integrate it directly. Instead, you can use the following trignometric identity: $sin^2x = \frac{1}{2}(1 – cos2x)$ This converts the $\sin^2x$ function into something you can work with.
  • $\int cos(\alpha\;x+\beta) \; dx = \frac{1}{\alpha} sin(\alpha \; x + \beta)$

 

Step-1

  • Let us use the information from the Toolbox above: $\int \; sin^2(2x + 5) \; dx = \frac{1}{2}(1 – cos2\;(2x+5) \; dx$.
  • This identity expands to $\frac{1}{2}(1 – cos(4x+10) \; dx$

Step-2

  • $\int \frac{1}{2} (1-cos(4x+10)) dx = \int \frac{1}{2}dx - \frac{1}{2} \int cos(4x+10)dx $
  • Now, $\int\frac{1}{2}\;dx = \frac{1}{2}$. We need to evaluate $\int cos(4x+10)\; dx$
Step-3
  • Hint: Remember: $\int cos(\alpha\;x+\beta) \; dx = \frac{1}{\alpha}sin(\alpha\;x+b)$
  • Therefore. $\int cos(4x+10)\; dx = \frac{1}{4}sin(4x+10)$

Step-4

  • Evaluating the entire term above, we get $\int \; sin^2(2x + 5) \; dx = \frac{1}{2}x - \frac{1}{2}\;\frac{1}{4}sin(4x+10) + k (Constant)$
  • Rewriting, we get the solution as $\frac{1}{2}x - \frac{1}{8}sin(4x+10) + k$

 

answered Dec 6, 2012 by balaji.thirumalai
edited Dec 6, 2012 by balaji.thirumalai
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...